Drawing the Lewis Structure for IF4 -

Viewing Notes:

  • In the IF4 - Lewis structure Iodine (I) is the least electronegative atom and goes in the center of the Lewis structure.
  • The IF4 - Lewis structure you'll need to put more than eight valence electrons on the Iodine atom.
  • In the Lewis structure for IF4 - there are a total of 36 valence electrons.

Transcript: This is the IF4- Lewis structure. For IF4-, we have a total of 36 valence electrons, which includes this valence electron here, as well. We'll put the Iodine at the center, it's the least electronegative, and then the Fluorines on the outside. We'll put a bond between each of the atoms. So we've used 2, 4, 6, 8. And then we'll fill the octets for the Fluorines on the outside. So we have 8, 10, 12, and 32. And we have 36 total valence electrons, but we've only used 32, so that leaves us with four more electrons. So we'll put a pair right here on the central Iodine, and let's put another pair right here.

So at this point, we've used all 36 valence electrons. The Fluorines all have an octet; and the Iodine, which is in period 5 on the periodic table, it's OK for it to have more than eight valence electrons, so it's fine, as well. If you check the formal charges, you'll see that the Fluorines have a formal charge, each one has a formal charge of zero. The Iodine has a negative one formal charge. That makes sense, because it's a negative ion. So that's fine, as well.

So this is the Lewis structure for IF4-. There's one last thing we need to do: we need to put brackets around the IF4- molecule here to show that it is an ion and has a negative charge. So we have our brackets, we have full octets for the Fluorines, and the Iodine's fine, as well. Our formal charges match up.

So this is the structure for IF4-, and this is Dr. B.. Thanks for watching.